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Solved Problem on Dynamics
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In the system of the figure, the body B slides on a horizontal surface without friction, it is connected through by ropes and pulleys, lightweight and frictionless, with two bodies A and C that move vertically. The masses of A, B, and C are, respectively, 5 kg, 2 kg and 3 kg. Find the acceleration of the system and the tension in the cord. Assume g = 10 m/s2.
Block B, of mass 2 kilograms, connected on both sides to block A, of mass 5 kilograms, and block C, of mass 3 kilograms, through pulleys.


Problem data
  • mass of body A:    mA = 5 kg;
  • mass of body B:    mB = 2 kg;
  • mass of body C:    mC = 3 kg;
  • free-fall acceleration:    g = 10 m/s2.
Problem diagram

We choose the acceleration in the direction in which the body A moves downward. Drawing a free-body diagram for each block and using Newton's Second Law
\[ \bbox[#99CCFF,10px] {\vec{F}=m\vec{a}} \]
Acceleration was chosen in the direction, that block A descends, block B moves from right to left and block C rises.
figure 1
Solution

Body A
  • \( \vec W_{\text{A}} \): weight of body A;
  • \( \vec T_{\text{AB}} \): tension in the cord between blocks A and B.
In the horizontal direction, there are no forces acting, in the vertical direction using Newton's Second Law we have
Block A with weight (PA) in the same direction as system acceleration (downward) and tension force (TAB) in the opposite direction.
figure 2
\[ P_{\text{A}}-T_{\text{AB}}=m_{\text{A}}a \tag{I} \]
Body B

Vertical direction
  • \( \vec W_{\text{B}} \): weight of body B;
  • \( \vec N_{\text{B}} \): normal force of the surface on the body.
Horizontal Direction
  • \( \vec T_{\text{AB}} \): tension in the rope between blocks A and B;
  • \( \vec T_{\text{BC}} \): tension in the rope between blocks B and C.
Block B act the weight (PB), normal reaction (NB), tension (TAB), in the same direction of acceleration, and tension (TBC) in the opposite direction.
figure 3
In the vertical direction, the weight and the normal force cancel out, there is no vertical motion.
In the horizontal direction by using Newton's Second Law we have
\[ T_{\text{AB}}-T_{\text{BC}}=m_{\text{B}}a \tag{II} \]
Corpo C
  • \( \vec W_{\text{C}} \): weight of body C;
  • \( \vec T_{\text{BC}} \): tension in the rope between blocks B and C.
In the horizontal direction, there are no forces acting, in the vertical direction by applying Newton's 2nd Law we have
Block C act the weight (PC), in the opposite direction of acceleration, and tension (TBC) in the same direction.
figure 4
\[ T_{\text{BC}}-W_{\text{C}}=m_{\text{C}}a \tag{III} \]
With equations (I), (II) and (III) we have a system of linear equations with three variables (TAB, TBC and a), adding the three equations we have
\[ \frac{ \left\{ \begin{array}{rr} W_{\text{A}}-\cancel{T_{\text{AB}}}&=m_{\text{A}}a\\ \cancel{T_{\text{AB}}}-\cancel{T_{\text{BC}}}&=m_{\text{B}}a\\ \cancel{T_{\text{BC}}}-W_{\text{C}}&=m_{\text{C}}a \end{array} \right.} {W_{\text{A}}-W_{\text{C}}=\left(m_{\text{A}}+m_{\text{B}}+m_{\text{C}}\right)a} \]
\[ a=\frac{W_{\text{A}}-W_{\text{C}}}{m_{\text{A}}+m_{\text{B}}+m_{\text{C}}} \tag{IV} \]
The weight of bodies A and C are given by
\[ W_{\text{A}}=m_{\text{A}}g\ \ \ \ \ \ \ \ \text{and}\ \ \ \ \ \ \ \ W_{\text{C}}=m_{\text{C}}g \tag{V} \]
substituting expression (V) in (IV) and the values given in the problem
\[ a=\frac{m_{\text{A}}g-m_{\text{C}}g}{m_{\text{A}}+m_{\text{B}}+m_{\text{C}}}\\ a=\frac{5 \times 10-3 \times 10}{5+2+3}\\ a=\frac{50-30}{10}\\ a=\frac{20}{10} \]
\[ \bbox[#FFCCCC,10px] {a=2\;\text{m/s}^{2}} \]
Substituting the mass of the body A, the expression for the weight of A given in (V) and the acceleration, found above, in the first expression of the system, the tension in the cord is
\[ m_{\text{A}}g-T_{\text{AB}}=m_{\text{A}}a\\ 5 \times 10-T_{\text{AB}}=5 \times 2\\ 50-T_{\text{AB}}=10\\ T_{\text{AB}}=50-10 \]
\[ \bbox[#FFCCCC,10px] {T_{\text{AB}}=40\ \text{N}} \]
Substituting the mass of the body C, the expression for the weight of C given in (V) and the acceleration, found above, in the third expression of the system, the tension in the cord is
\[ T_{\text{BC}}-m_{\text{C}}g=m_{\text{C}}a\\ T_{\text{BC}}-3.10=3.2\\ T_{\text{BC}}-30=6\\ T_{\text{BC}}=6+30 \]
\[ \bbox[#FFCCCC,10px] {T_{\text{AB}}=36\ \text{N}} \]

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